Proof that a function is continuous

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LASmith
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Prove that the function is continuous when f(x)=0
f(x)=x4-7x3+11x2+7x-12f(c)-[itex]\epsilon[/itex]<f(x)<f(c)+[itex]\epsilon[/itex]

Limits maybe taken, however, we do not have the value for c in the limit equation.
 
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It looks like you have to solve for x, when f(x) = 0, i.e., find the zeros of f(x).

For one of the zeros, notice that the sum of the coefficients of f(x) is zero. Therefore, f(1) = 0.

So you know that one of the factors of f(x) is (x-1). Use long division or synthetic division to find g(x) such that: f(x) = (x-1)g(x).

Added in Edit.
Notice that: f(-x) = x4+7x3+11x2-7x-12. Therefore, f(-(1)) = 0 .
 
Last edited:
SammyS said:
It looks like you have to solve for x, when f(x) = 0, i.e., find the zeros of f(x).

Solving this gives (x-1)(x+1)(x-4)(x-3)
But is this sufficient to show that it is continuous?
 
LASmith said:
Solving this gives (x-1)(x+1)(x-4)(x-3)
But is this sufficient to show that it is continuous?
Of course not.

The problem is to show that f(x) is continuous when f(x)=0. So the problem has become: show that f(x) is continuous for x = -1, 1, 3, 4 .